
Median of triangle
One of the medians of the triangle formed by the line pair $\displaystyle ax^2+2hxy+by^2=0$ and the line $\displaystyle px+qy=r$ lies along the yaxis. If neither a nor r is zero, prove that bp+hq=0.
The line pair has its vertex at (0,0), so that is the vertex of the median on the yaxis. Then the midpoint of the intersection of the line pair and the line would be the other point on the median and triangle.
When x=0, on the line, $\displaystyle y=\frac{r}{q}$
I found the sum of the points of intersection:
$\displaystyle x=\frac{rqy}{p}$
$\displaystyle a(\frac{rqy}{p})^2+2hy(\frac{rqy}{p})+by^2=0$
$\displaystyle y^2(aq^2+bp^22hpq)+y(2r(hpaq))+ar^2=0$
then
$\displaystyle \frac{1}{2}\times\frac{2r(hpaq)}{aq^2+bp^22hpq}=\frac{r}{q}$
works out to hqbp=0
I can't see where i have gone wrong.
Thanks

Your work out is correct.